Solution Set of Homework # 6 Monday, December 12, Textbook: Claude Cohen Tannoudji, Bernard Diu and Franck Laloë, Second Volume

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1 Department of Physics Quantum II, 570 Temple University Instructor: Z.-E. Meziani Solution Set of Homework # 6 Monday, December, 06 Textbook: Claude Cohen Tannoudji, Bernard Diu and Franck Laloë, Second Volume Complement D XIV Solution Exercise, page 447 We summarize the results of a) and b) in the tables below a) E g ω 0 ω ω ω ω 0 b) E g 0 ω 0 ω 0 3 ω 0 4 ω 0 5 ω 0 6 ω 0 a) h 0 is the Hamiltonian of a single particle, and acts only on the orbital variables. We are concerned with both the spin space and the subspace of E r, which is spanned by the three eigenstates of h 0 whose eigenvalues are 0, ω 0 and ω 0. These eigenvalues are non-degenerate in the orbital space of E r. Electrons are spin-/ particles, so the Pauli Exclusion Principle applies. The problem is to distribute the three electrons among three energy levels while satisfying the principle. Let and represent spin-up and spin-down electrons, respectively. Now, we draw all of the ways to fill the energy levels. Note: means up or down. Each contributes a factor of to the degeneracy. Energy Level ω 0 ω 0 0 Result E = E = E = E = E = E = E = ω 0 ω 0 ω 0 3 ω 0 4 ω 0 4 ω 0 5 ω 0 b) Same question for spin-0 bosons. Since they are bosons, Pauli Exclusion does not apply. Since they are spin-0, there is no degeneracy due to spin. Let (n) indicate the number of particles in each energy level: Energy Level ω 0 () () () () () (3) ω 0 () () () (3) () () 0 (3) () () () () () Result E = 0 E = E = E = E = E = E = E = E = E = ω 0 ω 0 ω 0 3 ω 0 3 ω 0 4 ω 0 4 ω 0 5 ω 0 6 ω 0 Exercise 6, page 449 Possible values of the relative angular momentum of two identical particles

2 a) First assume that the two particles are identical bosons of zero spin (π mesons for example). α. We use the { r G, r } basis of the state space E of the system, composed of common eigenvectors of the observables R G and R. We want to show that Since r = r r and r G = ( r + r ) we can write ˆP r G, r = r G, r () Now exchanging particle and we have ˆP = r G, r = ˆP r G = ( r + r ), r = ( r r ) () ˆP = r G, r = ˆP r G = ( r + r ), ( r r ) = ( r + r ), ( r r ) = r G = ( r + r ), r = ( r r ) = r G, r (3) β. We now go to the { p G ; E n, l, m } basis of common vectors of P G, H r, L and L z ( L = R P is the relative angular momentum of the two particles). We will show that these new basis vectors are given by expressions of the form: p G ; E n, l, m = R nl (r)yl m (θ, φ) r G, r (4) To show this we just need to insert the closure relation in the basis {vertr G, r } expressed as r G, r r G, r G = (5) p G ; E n, l, m = p G ; E n, l, m = r G, r r G, r p G ; E n, l, m G = G r G p G r E n, l, m r G, r (6) We have used the fact that r G, r = r G r where each ket is a state in a subspace of the center of mass space and the relative coordinate space. Now since we know that r G is a conjugate coordinate of p G, the projection coefficients from the position r G to the momentum eigenstates p G, are plane waves: r G p G = (π ) 3/ ei p G r G / Furthermore the projection of the Dirac states of the relative position representation gives r E n, l, m = R nl (r)yl m (θ, φ) (8) (7)

3 replacing these terms in the equation (6) above we get the desired result, namely p G ; E n, l, m = R nl (r)yl m (θ, φ) r G, r (9) Now we will show that: ˆP p G ; E n, l, m = ( ) l p G ; E n, l, m (0) If we act with the exchange operation on the equivalent expression of p G ; E n, l, m given by equation (9 ˆP p G ; E n, l, m = ˆP p G ; E n, l, m = r G, r r G, r ˆP p G ; E n, l, m G = r G, r r G, r p G ; E n, l, m G = G r G p G r E n, l, m r G, r = R nl (r)yl m (π θ, π + φ) r G, r = ( ) l R nl (r)yl m (θ, φ) r G, r = ( ) l p G ; E n, l, m () wehere we have used r G, r ˆP = r G, r and the Parity operator ˆ P. γ. It is clear that since the states describing this system of two bosons have to be symmetric, it is clear that l can only take even values to respect the symmetrization postulate. b) Now the two particles under consideration are identical fermions of spin / (electrons or protons). α. The state space of the system is { r G, r; S, M } basis of common eigenstates of R G, R, S and S z, where S = S + S is the total spin of the system. We want to show that ˆP r G, r; S, M = ( ) S+ r G, r; S, M () Since we can write the state vector r G, r; S, M as a tensor product of r G, r S, M of spatial states and spin states and because we have already seen the action of ˆP on the spatial states let us now determine its action on the spin states. From Section B of Chapter X we can write for the triplet states ˆP, + = ˆP : +; : + = : +; : + = : +; : + =, + ˆP, 0 = ˆP ( : +; : + : ; : + ) = ( : ; : + + : +; : ) =, 0 ˆP, = ˆP : ; : = : ; : = : ; : =, (3) 3

4 This result can be summarized in one equation ˆP S =, M = ( ) S=+ S =, M. Now for the singlet state we have ˆP 0, 0 = ˆP ( : +; : : ; : + ) = ( : ; : + : +; : ) = ( : +; : : ; : + ) = 0, 0 (4) which we can summarize as ˆP S = 0, 0 = ( ) S=0+ S =, M. Therefore combining the triplet and singlet state and the spatial part we arrive at the following result note the change of sign for the relative spatial variable r ˆP r G, r; S, M = ( ) S+ r G, r; S, M (5) β. Making the transformation from the r G, r; S, M states to the p G ; E n, l, m; SM states it is clear that we need to combine the result of a) β and b) α and can find through the tensor product of spin states and spatial states (expressed in terms of spherical coordinate system) the following result b β namely ˆP p G ; E n, l, m; S, M = ( ) S+ ( ) l p G ; E n, l, m; S, M (6) γ. The values allowed by the symmetrization for fermions are such that ˆP p G ; E n, l, m; S, M = p G ; E n, l, m; S, M (7) this implies that l + S + should be odd for all combinations of l and S thus l values should be odd for the triplet state and even values of l for the singlet state. c) The total scattering cross section in the center of mass of two distinguishable particles interacting through the potential V (r) is written as: σ = 4π k (l + ) sin δ l (8) with δ l as the phase shifts associated with V (r). l=0 α. When we have identical particles and the measurement device is equally sensitive to both particles it means that there is an ambiguity between two different scattering configurations which are indistinguishable, one where the angle of the scattered particle is θ and one where r r thus θ = π θ. Here the angles are described in the center of mass of the system of the two particles. β. The calculation of the cross section according to Chapter VIII formula C-56 needs to be augmented by a different scattering amplitude (f k (θ) + f k (π θ)) to include both possibilities of scattering. For the case of bosons f k (π θ) = ( ) l f k (θ) with thus f k (π θ) = +f k (θ). Thus the properly normalized scattering amplitude of formula C-55 should be with l is restricted to even values. This leads to (f k (θ) + f k (π θ)) = (f k (θ) + f k (θ)) σ = f k (θ) (9) σ = 8π k (l + ) sin δ l (0) γ. Note that the sum is only on the even values of the quantum number l. For the case of fermions f k (π θ) = ( ) l f k (θ) but here we need to include the spin of the particles. Therefore for the triplet spin 4

5 states l is odd with f k (π θ) = f k (θ) and for the singlet case l is even leading to f k (π θ) = f k (θ). Thus the scattering amplitude of formula is (f k (θ) f k (π θ)) = (f k (θ) + f k (θ)) = f k (θ) = f k (θ) () In order to evaluate the identical fermions cross section we need to take into account that the total spin part of the amplitude is normalized properly. For the cross section we have /4 probability that the scattering happens when the two particles are in a singlet state configuration with even l values and 3/4 probability that the scattering occurs when the particles are in a triplet state configuration with l is restricted to even values. This leads to σ = 4π k { σ = π k /4 (l + ) sin δ l + 3/4 } (l + ) sin δ l l odd { (l + ) sin δ l + 3 } (l + ) sin δ l l odd () 5